An aeroplane flies at a height h = 16 km with a body of mass…

An aeroplane flies at a height $h = 16\ \mathrm{km}$ with a body of mass $m = 1\ \mathrm{kg}$ on board. The Earth's mean radius is $R = 6374\ \mathrm{km}$ and its mass $M = 6 \cdot 10^{24}\ \mathrm{kg}$. (a) What gravitational force acts on the body? (b) What is the weight of the body if the aeroplane flies at constant velocity over the pole, or over the equator?

Step 1: Idea

We must distinguish two things. The gravitational force is the Earth's pull, given by Newton's law; it depends only on the distance from the centre. The weight is the force with which the body presses on its support (the floor of the plane), which is what a scale would show. Because the Earth rotates, the body moves on a circle around the axis, and part of the gravitational force is used up to keep it on that circle. So the weight is slightly smaller than the gravitational force, except where there is no circular motion.

Step 2: (a) Gravitational force

The distance from the Earth's centre is $R + h = 6390\ \mathrm{km}$: \[F = \gamma\frac{Mm}{(R + h)^2} = \frac{6.67 \cdot 10^{-11} \cdot 6 \cdot 10^{24} \cdot 1}{(6.39 \cdot 10^6)^2}.\]

\[F \approx 9.80\ \mathrm{N}\]

At the surface it would be $9.85\ \mathrm{N}$ for these data; 16 km of height reduces it by only 0.5%.

Step 3: (b) Over the pole

A body above the pole lies on the Earth's axis. It turns on the spot but does not move on a circle, so no centripetal force is needed and the whole gravitational force presses the body against the floor:

\[W_{\text{pole}} = F \approx 9.80\ \mathrm{N}\]

Step 4: (b) Over the equator

Here the body goes round the axis once a day on a circle of radius $R + h$. The centripetal force this requires is

\[ m\omega^2(R + h) = 1 \cdot (7.27 \cdot 10^{-5})^2 \cdot 6.39 \cdot 10^6, \]

\[ m\omega^2(R + h) = 0.034\ \mathrm{N}. \]

Gravity must supply this, so only the remainder presses on the floor: $W = F - m\omega^2(R + h) = 9.801 - 0.034$.

\[W_{\text{eq}} \approx 9.77\ \mathrm{N}\]

The body weighs about 0.35% less over the equator. (We have ignored the plane's own speed over the ground; flying east would slightly increase the effect, flying west slightly reduce it.)

Check

The centrifugal part is $0.034/9.80 = 0.35\%$ of the pull, the known difference between equatorial and polar weight due to rotation $\checkmark$